Here, we will prove that the process of decomposing is unique, given that M is noetherian and artinian.

Again, R is a ring, possibly non-commutative.

Definition. A decomposition of an R-module M is an expression for non-zero modules

An R-module M is said to be indecomposable if no such expression exists for k≥2 (or equivalently, for k=2).

A simple module is clearly indecomposable, but the converse is not true.

For example, let be the ring of upper-triangular 2 × 2 real matrices. Consider the module M of column vectors R2, where R acts by matrix multiplication. Then M is not simple since it has a submodule spanned by (1, 0). But M is indecomposable since any submodule N ⊂ M which contains an (x, y) for y≠0 must be the whole M.

We have a criterion for a module to be indecomposable:

Proposition. Let M be an artinian and noetherian R-module. Then M is indecomposable if and only if is a local ring.

⇒ : recall that for an artinian and noetherian M, a module map f : M → M is bijective iff it is surjective or injective. Now suppose f, g : M → M are module maps such that f+g is a unit and g is not. We need to show that f is a unit.

Write (f+g)h = 1 for some h : M → M, so fh = 1-gh. Now k := gh is not an isomorphism since g is not surjective. Since M is artinian and noetherian, eventually

for some large n. We claim that

To show : for any x∈M, we have so we can write But then ; clearly the second term lies in and the first term gives so it lies in

To show : if lies in then we have which gives and so

Since M is indecomposable either or The former case says k is injective and hence an isomorphism, so we have the latter case kn=0. Thus fh = 1-k is invertible (since 1 + nilpotent = unit), so f is right-invertible. By symmetry, it is also left-invertible. ♦

The Krull-Schmidt Theorem

Before stating and proving our main theorem, here is a useful criterion for splitting a module as a direct sum:

Splitting Lemma. If satisfy then N splits as Furthermore f is injective, thus M is a direct summand of N.

Krull-Schmidt Theorem. Let M be an artinian and noetherian module. If we decompose:

then k=l and there is a permutation σ of {1, 2, …, k} such that for all i=1,…,k.

Proof

We shall prove that U1 ≅ Vj for some j, such that after which we can apply induction on k.

Let be the projection onto U1 and be the projection onto Vi. This gives and thus Since im(p) ⊆ U1 this restricts to a map

Since U1 is indecomposable, End(U1) is local; and since , one of the αj is invertible, say α1. Let be the inverse of , so

where (*)

From the splitting lemma, U1 is a direct summand of V1; since V1 is indecomposable we must have U1≅V1. To prove that we write (*) as

where

Again, splitting lemma tells us The second term is isomorphic to V1 and the first is just ker(p), which is as desired. ♦

Note

Because M is artinian and noetherian it is always possible to decompose M as a direct sum of finitely many indecomposable modules. Simply keep decomposing the terms until we’re left with indecomposable modules: since M is of finite length, the process must terminate. Krull-Schmidt theorem says that the resulting terms are unique to isomorphism and permutation.

Example

Let R be the ring of upper-triangular 2 × 2 matrices with real entries. Here are two different ways of decomposing R as a direct sum of indecomposable left ideals:

Regardless of the decomposition we pick, the second term must be there (prove this!).